Questions: Compare the monthly payment and total payment for the following pairs of loan options. Assume that both loans are fixed rate and have the same closing costs. You need a 110,000 loan. Option 1: a 30-year loan at an APR of 8%. Option 2: a 15-year loan at an APR of 7% Find the monthly payment for each option. The monthly payment for option 1 is The monthly payment for option 2 is Find the total payment for each option The total payment for option 1 is The total payment for option 2 is Compare the two options. Which appears to be the better option? Option 1 is the better option, but only if the borrower plans to stay in the same home for the entire term of the loan. B. Option 2 will always be the better option. Option 1 will always be the better option Option 2 is the better option, but only if the borrower can afford the higher monthly payments over the entire term of the loan

Solution Steps

To solve this problem, we need to calculate the monthly payment for each loan option using the formula for monthly payments on an amortizing loan. The formula is:

\[ M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1} \]

where:

• \( M \) is the monthly payment,
• \( P \) is the principal loan amount (\$110,000 in this case),
• \( r \) is the monthly interest rate (annual rate divided by 12),
• \( n \) is the total number of payments (loan term in years multiplied by 12).

After calculating the monthly payments, we can find the total payment by multiplying the monthly payment by the total number of payments. Finally, we compare the total payments and monthly payments to determine which option is better based on the given conditions.

For Option 1, a 30-year loan at an APR of \(8\%\):

• Principal \(P = 110,000\)
• Monthly interest rate \(r_1 = \frac{0.08}{12} = 0.0066667\)
• Total payments \(n_1 = 30 \times 12 = 360\)

The monthly payment \(M_1\) is calculated using the formula: \[ M_1 = \frac{P \times r_1 \times (1 + r_1)^{n_1}}{(1 + r_1)^{n_1} - 1} \] Substituting the values: \[ M_1 = \frac{110,000 \times 0.0066667 \times (1 + 0.0066667)^{360}}{(1 + 0.0066667)^{360} - 1} \approx 807.14 \]

The total payment \(TP_1\) for Option 1 is: \[ TP_1 = M_1 \times n_1 = 807.14 \times 360 \approx 290,570.77 \]

For Option 2, a 15-year loan at an APR of \(7\%\):

• Monthly interest rate \(r_2 = \frac{0.07}{12} = 0.0058333\)
• Total payments \(n_2 = 15 \times 12 = 180\)

The monthly payment \(M_2\) is calculated using the same formula: \[ M_2 = \frac{P \times r_2 \times (1 + r_2)^{n_2}}{(1 + r_2)^{n_2} - 1} \] Substituting the values: \[ M_2 = \frac{110,000 \times 0.0058333 \times (1 + 0.0058333)^{180}}{(1 + 0.0058333)^{180} - 1} \approx 988.71 \]

The total payment \(TP_2\) for Option 2 is: \[ TP_2 = M_2 \times n_2 = 988.71 \times 180 \approx 177,968.00 \]

• Monthly payment for Option 1: \(M_1 \approx 807.14\)
• Total payment for Option 1: \(TP_1 \approx 290,570.77\)
• Monthly payment for Option 2: \(M_2 \approx 988.71\)
• Total payment for Option 2: \(TP_2 \approx 177,968.00\)

Since \(TP_2 < TP_1\), Option 2 is the better option if the borrower can afford the higher monthly payments over the entire term of the loan.

Final Answer